### Local Leopoldt’s Problem

### for Rings of Integers in Abelian

p### -Extensions of Complete Discrete Valuation Fields

M. V. Bondarko

Received: April 12, 2000 Revised: December 8, 2000

Communicated by A. S. Merkurev

Abstract. Using the standard duality we construct a linear embedding of an associated module for a pair of ideals in an extension of a Dedekind ring into a tensor square of its fraction field. Using this map we investi- gate properties of the coefficient-wise multiplication on associated orders and modules of ideals. This technique allows to study the question of de- termining when the ring of integers is free over its associated order. We answer this question for an Abelian totally wildly ramified p-extension of complete discrete valuation fields whose different is generated by an element of the base field. We also determine when the ring of integers is free over a Hopf order as a Galois module.

1991 Mathematics Subject Classification: 11S15, 11S20, 11S31

Keywords and Phrases: Complete discrete valuation (local) fields, addi- tive Galois modules, formal groups, associated Galois modules

Introduction

0.1. Additive Galois modules and especially the ring of integers of local fields
are considered from different viewpoints. Starting from H. Leopoldt [L] the
ring of integers is studied as a module over its associated order. To be precise,
if K is an extension of a local field k with Galois group being equal to G
and O_{K} is the ring of integers ofK, thenO_{K} is considered as a module over
A_{K/k}(O_{K}) ={λ∈k[G], λO_{K} ⊂O_{K}}.

One of the main questions is to determine whenO_{K} is free as anA_{K/k}(O_{K})-
module. Another related problem is to describe explicitly the ringA_{K/k}(O_{K})
(cf. [Fr], [Chi], [CM]).

This question was actively studied by F. Bertrandias and M.-J. Ferton (cf. [Be], [B-F], [F1-2]) and more recently by M. J. Taylor, N. Byott and G. Lettl (cf.

[T1], [By], [Le1]).

In particular, G. Lettl proved that ifK/Q_{p} is Abelian andk⊂K, thenO_{K} ≈
A_{K/k}(O_{K}) (cf. [Le1]). The proof was based on the fact that all Abelian
extensions ofQ_{p} are cyclotomic. So the methods of that paper and of most of
preceding ones are scarcely applicable in more general situations.

M. J. Taylor [T1] considers intermediate extensions in the tower of Lubin-Tate
extensions. He proves that for some of these extensionsO_{K}is a freeA_{K/k}(O_{K})-
module. Taylor considers a formal Lubin-Tate groupF(X, Y) over the ring of
integerso of a local fieldk. Let π be a prime element of the field k and Tm

be equal to Ker[π^{m}] in the algebraic closure of the fieldk. For 1≤r≤m, let
L be equal to k(Tm+r), and let K be equal to k(Tm), . Lastly, let q be the
cardinality of the residue field ofk.

Taylor proves that

(1) the ringO_{L} is a freeA_{L/K}(O_{L})-module and any element ofLwhose valu-
ation is equal toq^{r}−1 generates it, and

(2) A_{L/K}(O_{L}) = O_{K} +Pq^{r}−2

i=0 O_{K}σi, where σi ∈ K[G] and are described
explicitly (cf. the details in [T1], subsection 1.4).

This result was generalized to relative formal Lubin-Tate groups in the papers [Ch] and [Im]. Results of these papers were proved by direct computation. So these works do not show how one can obtain a converse result, i.e., how to find all extensions, that fulfill some conditions on the Galois structure of the ring of integers. To the best of author’s knowledge the only result obtained in this direction was proved in [By1] and refers only to cyclotomic Lubin-Tate extensions.

0.2. In the examples mentioned above the associated order is also a Hopf order
in the group algebra (i.e., it is an order stable under comultiplication). Sev-
eral authors are interested in this situation. The extra structure allows to deal
with wild extensions as if they were tame (in some sense). This is why in this
situation, following Childs, one speaks of “taming wild extensions by Hopf or-
ders”. In the paper [By1], Byott proves that the associated order can be a Hopf
order only in the case when the different of the extension is generated by an
element of the smaller field. The present paper is also dedicated to extensions
of this sort. More precisely, Theorem 4.4 of the paper [By1] implies that the
order A_{K/k}(O_{K}) can be a Hopf order (in the case when the ringO_{K} iso[G]-
indecomposable) only if K/k fulfills the stated condition on the different and
O_{K}is free overA_{K/k}(O_{K}). Our main Theorem settles completely the question
to determine when the orderA_{K/k}(O_{K}) is a Hopf order. It also describes com-
pletely Hopf orders that can be obtained as associated Galois orders. We shall
also prove in subsection 3.4 that under the present assumptions ifO_{K} is free
over A_{K/k}(O_{K}), then A_{K/k}(O_{K}) is a Hopf order and determine when the in-
verse different of an Abelian totally ramifiedp-extension of a complete discrete
valuation field is free over its associated order (cf. [By1] Theorem 3.10).

In the first section we study a more general situation. We consider a Galois extension K/k of fraction fields of Dedekind rings, with Galois group G. We prove a formula for the module

C_{K/k}(I1, I2) = Homo(I1, I2),

where I1, I2 are fractional ideals of K (this set also can be defined for ideals
that are not G-stable) in Theorem 1.3.1. We introduce two submodules of
C_{K/k}(I1, I2):

A_{K/k}(I1, I2) ={f ∈k[G]|f(I1)⊂I2}
B_{K/k}(I1, I2) = Homo[G](I1, I2), .

These modules coincide in the case whenK/kis Abelian (cf. Proposition 1.4.2).

We call these modules theassociated modulesfor the pairI1, I2. In caseI1=I2

we call the associated moduleassociated order.

In subsection 1.5 we define a multiplication on the modules of the type
C_{K/k}(I1, I2) and show the product of two such modules lies in some third one.

We have also used this multiplication in the study of the decomposability of ideals in extensions of complete discrete valuation fields with inseparable residue field extension (cf. [BV]).

Starting from the second section we consider totally wildly ramified extensions of complete discrete valuation fields with residue field of characteristicpwith the restriction on the different:

D_{K/k} = (δ), δ∈k (∗).

This second section is dedicated to the study of conditions ensuring that the
ring of integersO_{K} is free over its associated orderA_{K/k}(O_{K}) orB_{K/k}(O_{K}).

Letn= [K:k].

We prove the following statement.

Proposition. If in the associated order A_{K/k}(O_{K})(B_{K/k}(O_{K}) resp.) there
is an elementξ which maps some (and so, any) elementa∈O_{K} with valuation
equal to n−1onto a prime element of the ring O_{K}, then O_{K} ≈A_{K/k} (resp.

B_{K/k}) and besides thatA_{K/k} (resp. B_{K/k}) has a “power” base (in the sense of
multiplication ^{∆}∗, cf. the second section), which is constructed explicitly using
the elementξ.

The converse to this statement is also proved in the case whenA_{K/k}(O_{K}) (resp.

B_{K/k}(O_{K})) is indecomposable (cf. the Theorems 2.4.1, 2.4.2). An important
part of our reasoning is due to Byott (cf. [By1]).

0.3. The third, fourth and fifth sections are dedicated to proving a more ex- plicit form of the condition of the second section for the Abelian case. The main result of the paper is the following one. We will call an Abelianp-extension of complete discrete valuation fields of characteristic 0almost maximally ramified if its degree divides the different.

Theorem A. Let K/k be an Abelian totally ramified p-extension of p-adic fields, which in case chark= 0 is not almost maximally ramified, and suppose that the different of the extension is generated by an element in the base field (see (*)). Then the following conditions are equivalent:

1. The extension K/k is Kummer for a formal group F, that there exists
a formal group F over the ring of integers o of the field k, a finite torsion
subgroupT of the formal moduleF(M_{o})and a prime elementπ0ofksuch that
K=k(x), wherex is a root of the equationP(X) =π0, where

P(X) =Y

t∈T

(X−

F t).

2. The ring O_{K} is isomorphic to the associated order A_{K/k}(O_{K})as an o[G]-
module.

Remark 0.3.1. Besides proving Theorem A we will also construct the element
ξ explicitly and so describeA_{K/k}(O_{K}) (cf. the theorems of §2).

Remark 0.3.2. IFkis of characteristic 0 the fact thatO_{K}is indecomposable as
ano[G]-module if and only ifK/kis not almost maximally ramified, was proved
in [BVZ]. The case of almost maximally ramified extensions is well understood
(cf., for example, [Be]). It is obvious, that in the case chark=pthe ringO_{K}
is indecomposable as ano[G]-module, because in this case the algebrak[G] is
indecomposable.

Remark 0.3.3. In the paper [CM] rings of integers in Kummer extensions for formal groups are also studied as modules over their associated orders. In that paper Kummer extensions are defined with the use of homomorphisms of formal groups. For extensions obtained in this way freeness of the ring of integers over its associated orders is proved. Childs and Moss also use some tensor product to prove their results. Yet their methods seem to be inapplicable for proving inverse results.

Our notion of a Kummer extension for a formal group is essentially equivalent to the one in [CM]. We however only use one formal group and do not impose finiteness restriction on its height.

Besides we consider also the equal characteristic case i.e., chark= chark=p.

Using the methods presented here one can prove that we can actually take the formal groupF in Theorem A of finite height. Yet such a restriction does not seem to be natural.

Remark 0.3.4. Theorem A shows which Hopf orders can be associated to Galois orders for some Abelian extensions. The papers of mathematicians that “tame wild extensions by Hopf orders” do not show that their authors know or guess that such an assertion is valid.

In the third section we study the fields that are Kummer in the sense of part
1 of Theorem A and deduce 2 from 1. In the fourth section we prove that we
can suppose the coefficient b1 in ξ =δ^{−1}P

bσσ to be equal to 0. Further, if

b1is equal to 0, then we show that there exists a formal groupF overo, such
that forσ ∈Gthebσ, form a torsion subgroup in the formal module F(M_{o}),
i.e.,

bσ+

F bτ =bστ . In the fifth section we prove that ifbσ+

Fbτ is indeed equal tobστ, thenK/kis Kummer for the groupF.

0.4. This paper is the first in a series of papers devoted to associated orders and associated modules. The technique introduced in this paper (especially the mapφand the multiplication∗defined below) turns out to be very useful in studying The Galois structure of ideals. It allows the author to prove in another paper some results about freeness of ideals over their associated orders in extensions that do not fulfill the condition on the different (*). In particular, using Kummer extensions for formal groups we construct a wide variety of extensions in which some ideals are free over their associated orders. These examples are completely new. We also calculate explicitly the Galois structure of all ideals in such extensions. Such a result is very rare. In a large number of cases the necessary and sufficient condition for an ideal to be free over its associated order is found.

The author is deeply grateful to professor S. V. Vostokov for his help and advice.

The work paper is supported by the Russian Fundamental Research Foundation N 01-00-000140.

§1 General results Let

obe a Dedekind ring,

kbe the fraction field of the ring o,

K/k be a Galois extension with Galois group equal toG,
D=D_{K/k} be the different of the extensionK/k,

n= [K:k],

tr be the trace operator inK/k.

We also define some associated modules.

ForI1 andI2 G-stable ideals in Kput

B_{K/k}(I1, I2) = Hom^{o}[G](I1, I2). ForI1, I2not beingG-stable one can define
B_{K/k}(I1, I2) ={f ∈Homo[G](K, K), f(I1)⊂I2}.

For arbitraryI1, I2⊂K we define
A_{K/k}(I1, I2) ={f ∈k[G]|f(I1)⊂I2},
C_{K/k}(I1, I2) = Hom^{o}(I1, I2).

Obviously, any o-linear map from I1 into I2 can be extended to a k-linear
homomorphism fromKintoK. The dimension of Homk(K, K) overkis equal
to n^{2}. Now we consider the group algebraK[G]. This algebra acts onK and

the statement in (Bourbaki, algebra,§7, no. 5) implies that a non-zero element
ofK[G] corresponds to a non-zero map fromKintoK. Besides the dimension
ofK[G] overkis also equal ton^{2}. It follows that any element of Homk(K, K)
can be expressed uniquely as an element of K[G]. So we reckon C_{K/k}(I1, I2)
being embedded inK[G].

1.1. We consider theG-Galois algebraK⊗kK. It is easily seen that the tensor productK⊗kK is isomorphic to a direct sum ofncopies ofKas ak-algebra.

Being more precise, letKσ, σ∈Gdenote a field, isomorphic toK.

Lemma 1.1.1. There is an isomorphism of K-algebras ψ: K⊗kK→ X

σ∈G

Kσ,

whereψ=P

σψσ andψσ is the projection on the coordinate σ, defined by the equality

ψσ(x⊗y) =xσ(y)∈Kσ.

The proof is quite easy, you can find it, for example, in ”Algebra” of Bourbaki.

Also see 1.2 below.

Now we construct a mapφfrom theG-Galois algebraK⊗kK into the group algebraK[G]:

(1)

φ:K⊗kK→K[G]

α=X

i

xi⊗yi→φ(α) =X

i

xi

X

σ∈G

σ(yi)σ

! .

It is clear thatφ(α) as a function acts onK as follows:

(2) φ(α)(z) =X

i

xitr(yiz), z∈K.

Besides that, the mapφmay be expressed throughψσ in the form

(3) φ(α) =X

σ

ψσ(α)σ, α∈K⊗kK.

Proposition 1.1.2. The mapφis an isomorphism ofk-vector spaces between K⊗kK andK[G].

Cf. the sketch of the proof in Remark 1.2 below.

1.2 Pairings on K⊗kK and K[G].

There is a natural isomorphism of the k-space K and its dual space of k- functionals:

K→Kb a→fa(b) = tr(ab).

We define a pairingh,i⊗ onK⊗kK, that takes its values ink:

(K⊗kK)×(K⊗kK)→k

ha⊗b, c⊗di⊗ →(fa⊗fc)(b⊗d) = (trac)(trbd).

This pairing is correctly defined and is non-degenerate. The second fact is easily proved with the use of dual bases.

We also define a pairing onK[G] that takes its values ink:

K[G]×K[G]→k α=X

σ

aσσ, β=X

σ

bσσ→X

σ

traσbσ=hα, βiK[G].

The pairing h,iK[G] is also non-degenerate because it is a direct sum of n= [K:k] non-degenerate pairings

K×K→k (a, b)→trab.

We check that for any twox, y the following equality is fulfilled:

(4) hx, yi⊗=hφ(x), φ(y)iK[G].

Indeed, it follows from linearity that it is sufficient to prove that for x = a⊗b, y=c⊗d. In that case we have

ha⊗b, c⊗di⊗= tractrbd, and besides that

hφ(a⊗b), φ(c⊗d)iK[G]

=haX

σ

σ(b)σ, cX

σ

σ(d)σiK[G]

=X

σ

tr(acσ(bd)) =X

σ

X

τ

τ(acσ(bd))

=X

τ

X

σ

τ(ac)τ σ(bd) = tractrbd.

and the equality (4) is proved.

Remark 1.2. It follows from (4) that the mapφfrom (1) is an injection. Indeed, let φ(α) be equal to 0 for α ∈ K⊗kK, then hφ(α), φ(β)iK[G] = 0 for any b ∈ K⊗kK. So, according to (4), hα, βi⊗ = 0 for any β ∈ K⊗kK. From the non-degeneracy of the pairing h,i⊗ it follows that α = 0. That implies Ker(φ) = 0.

Using the equality of dimensions we can deduce that φis an isomorphism.

1.3 Modules of homomorphisms for a pair of ideals.

LetI1, I2 be fractional ideals of the fieldK.

Theorem 1.3.1. Letφbe the bijection from (1) and letI_{1}^{∗}=D^{−1}I_{1}^{−1} (this is
dual of I1 for the bilinear trace form on K). For the associated modules the
following equality holds:

C_{K/k}(I1, I2) =φ(I2⊗oI_{1}^{∗}).

Proof. First we show thatφ(I2⊗oI_{1}^{∗})⊂C_{K/k}(I1, I2). If x∈I2, y∈I_{1}^{∗}, then
for anyz∈I1we have, according to the definition (2):

φ(x⊗y)(z) =xtr(yz)∈I2,

sincex∈I2 and tr(yz)∈o, which follows from the definition ofI_{1}^{∗} and of the
differentD.

Conversely, let f ∈C_{K/k}(I1, I2). We define a mapθf and show that it is sent
ontof byφ. Let:

(5) θf : I_{2}^{∗}⊗^{o}I1→o

x⊗y→tr(xf(y)).

This map is correctly defined sincex∈I_{2}^{∗}=D^{−1}I_{2}^{−1} andf(y)∈I2, thus
tr(xf(y))∈o.

For ao-moduleM we denote byMcthe module ofo-linear functions from M into o. It is clear that

(6) θf ∈(I_{2}^{∗}\⊗^{o}I1).

We identify the idealI2 with theo-moduleIb_{2}^{∗} via:

I2→Ib_{2}^{∗}
a→fa =X

σ∈G

σ(a)σ.

Obviouslyfa(z) = tr(az) for anyz∈Ib2.

In a completely analogous way we identifyI_{1}^{∗} withIb_{1}^{∗}

Using these identifications and the fact thatI1andI_{2}^{∗}are projectiveo-modules
we obtain an isomorphism

(7) I2⊗^{o}I_{1}^{∗}→(I_{2}^{∗}\⊗^{o}I1)
a⊗b→ha,b,

whereha,b(x⊗y) =ha⊗b, x⊗yi⊗= (trax)(trby) for all xin I_{2}^{∗} andy inI1.
The mapθf in (5) lies in(I_{2}^{∗}\⊗^{o}I1) (cf. (6)), and so, according to the isomor-
phism (7), it corresponds to an element αf inI2⊗^{o}I_{1}^{∗}, i.e.,

αf =X

i

ai⊗bi, ai∈I2, bi∈I_{1}^{∗}.

Then (7) implies that the functionalhα_{f}, that corresponds to the elementαf,
is defined in the following way:

hαf(x⊗y) =hαf, x⊗yi⊗=X

i

traixtrbiy.

On the other hand, from the definition ofθf (cf. (5)) we obtain:

hα_{f}(x⊗y) =θf(x⊗y) = tr(xf(y)).

It follows thatf(y) =P

iaitr(biy). So we have f =φ(X

ai⊗bi),

and for anyf in C_{K/k}(I1, I2) we have found its preimage in I2⊗I_{1}^{∗}, i.e.,
C_{K/k}(I1, I2)⊂φ(I2⊗^{o}I_{1}^{∗}).

The theorem is proved.

Remark 1.3.2. In the same way as above we can prove, that if we replace the idealsI1, I2 by two arbitrary free o-submodules X and Y ofK of dimension n, then we will obtain the following formula:

C_{K/k}(X, Y) =φ(Y ⊗X),b

whereXb is the dual module toX inKwith respect to the pairingK×K→k defined by the trace tr.

Remark 1.3.3. All elements in C_{K/k}(I1, I2), A_{K/k}(I1, I2), B_{K/k}(I1, I2) have
unique extensions to k-linear maps from K to K. To be more precise, if f :
I1→I2is ano-homomorphism, then for allx∈Kwe can assumef(x) =αf(a)
ifx=αa, α∈k, a∈I1.It is easily seen that the map we obtained in this way
is a correctly definedk-linear homomorphism from KintoK.

1.4. Now we compare the modulesAandB. Proposition 1.4.

(8) A_{K/k}(I1, I2) =B_{K/k}(I1, I2)
if and only if K/k is an Abelian extension.

Proof. 1. LetK/k be an Abelian extension. To verify the equality (8) in this
case, let firstf belong to A_{K/k}(I1, I2), then f is ano-homomorphism fromI1

intoI2. Besides that,f commutes with all elements ofGsinceGis an Abelian
group, i.e., σf(a) =f(σ(a)) for all σ ∈Ganda∈I1. So we obtain that f is
ano[G]-homomorphism fromI1into I2, thusf ∈B_{K/k}(I1, I2).

For the reverse inclusion, let f belong to B_{K/k}(I1, I2). Then f induces an
o[G]-homomorphism from K into K. We take an element x that generates a
normal base of the fieldK overk. Then there exists an elementg∈k[G] such
that f(x) =g(x), to be more precise, if f(x) = P

σaσσ(x), aσ ∈k then we takeg=P

aσσ. We consider theo[G]-homomorphismgfromKintoK. Since Gis an Abelian group,f(σ(x)) =σ(f(x)) =σ(g(x)) =g(σ(x)) for anyσ∈G.

We obtain thatk-homomorphismsf andgcoincide on the basic elements and
so f =g∈k[G] and f(I1)⊂I2, i.e.,f ∈A_{K/k}(I1, I2).

2. Now we suppose thatA_{K/k}(I1, I2) =B_{K/k}(I1, I2) and check thatK/kis an
Abelian extension. Indeed, since kA_{K/k}(I1, I2) = k[G], A_{K/k}(I1, I2) contains
elements of the form aσ, where a ∈ k^{∗}, for any σ ∈ G. It follows from our
assumption that aσ∈B_{K/k}(I1, I2), and soaσ is ano[G]-homomorphism. We
obtain thatGcommutes with all elementsσ∈G. Proposition is proved.

Proposition 1.5. If we assume the action of Gon K⊗kK to be diagonal, then

B_{K/k}(I1, I2) =φ((I2⊗oI_{1}^{∗})^{G}).

Proof. Let α belong to (I2⊗^{o}I_{1}^{∗})^{G}. We have to show that φ(a) is an o[G]-
homomorphism fromI1into I2. This means that

σφ(a)(z) =φ(a)(σ(z)) for all z ∈I1. Let αbe equal to P

ai⊗bi, ai ∈I2, bi ∈I_{1}^{∗}. Then from the
definition of φ(cf. (2)) it follows that

φ(a)(σ(z)) =X

i

aitr(biσ(z))

=X

i

aitr(σ^{−1}(bi)σ(z)) =φ X

i

ai⊗σ^{−1}(bi)

! (z).

On the other hand,

σφ(a)(z) =σX

aitr(biz)

=X

(σ(ai) tr(biz)) =φX

σ(ai)⊗bi

(z).

From theG-invariance of the elementαit follows that

φX

ai⊗σ^{−1}(bi)

=φX

σ(ai)⊗σ(σ^{−1}(bi))

=φX

σ(ai)⊗bi

.

Thusφ(α)(σ(z)) =σφ(α)(z) for anyσ ∈G, i.e., φ(α)∈B_{K/k}(I1, I2).

Conversely, let f belong to B_{K/k}(I1, I2), then f ∈ C_{K/k}(I1, I2) and so, ac-
cording to Theorem 1.3.1, there is an α ∈ I2⊗I_{1}^{∗}, such that f = φ(α). It
remains to check that α is G-invariant. We use the fact that f is an G-
homomorphism, i.e., σf(z) = f(σz) for all σ ∈ G and z ∈ I1. We obtain
an equalityσφ(α)(z) =φ(α)(σz). By writing the left and the right side of the
equality as above we obtain forα=P

ai⊗bi: σφ(α)(z) =φX

σai⊗σ(σ^{−1}bi)
(z)
φ(α)(σz) =φX

ai⊗σ^{−1}bi

(z).

It follows that

φX

σai⊗σ(σ^{−1}bi)

=φX

ai⊗σ^{−1}bi

.

Now using the fact thatφis a bijection we obtain
Xσai⊗σ(σ^{−1}bi) =X

ai⊗σ^{−1}bi.
We apply to both sides of the equality the map

1⊗σ: K⊗kK→K⊗kK a⊗b→a⊗σ(b) , that obviously is an homomorphism. We have

Xσai⊗σbi=X ai⊗bi, i.e.,σ(α) =α.

1.6 The multiplication ∗onK[G].

On the algebraK⊗kKthere is a natural multiplication: (a⊗b)·(c⊗d) =ac⊗bd.

Using it and the bijection φwe define a multiplication on K[G]. To be more precise, iff, g∈K[G], then we define

f ∗g=φ(φ^{−1}(f)·φ^{−1}(g))∈K[G]. (9)

Proposition 1.6.1. Iff =P

σaσσ and g=P

σbσσ, then f∗g=X

σ

aσbσσ.

Proof. Let f be equal to φ(α), g be equal to φ(β), where α, β ∈K⊗kK. If α=Pxi⊗yi, β=Puj⊗vj, then from the definition ofφ(cf. (1)) we obtain

φ(α) =X xi

X

σ

σyiσ, φ(β) =X uj

X

σ

σvjσ.

It follows that X

σ

aσbσσ =X

i,j

xiujσ(yivj)σ.

On the other hand,

f ∗g=φ(αβ) =φ

X

i,j

(xiui⊗yivj)

=X

i,j

xiujσ(yivj)σ

and we obtain the proof of Proposition.

Remark 1.6.2. The formula from Proposition 1.6.1 will be used further as an another definition of the multiplication∗.

1.7 Multiplication on associated modules.

Now we consider the multiplication (9) on the different associated modules.

Here we will see appearing the different which we will suppose to be induced from the base field in the following sections.

Proposition 1.7.1. Let f belong to C_{K/k}(I1, I2), and let g belong to
C_{K/k}(I3, I4). Then

f ∗g∈C_{K/k}(I1I3D, I2I4).

Proof. It is clear that φ^{−1}(f) and φ^{−1}(g) belong to I2⊗oI_{1}^{∗} and I4⊗I_{3}^{∗} re-
spectively. So we obtain thatφ^{−1}(f)φ^{−1}(g) lies in the product

(I2⊗oI_{1}^{∗})(I4⊗oI3∗) =I2I4⊗o(I_{1}^{∗}I_{3}^{∗})

=I2I4⊗oD^{−2}I_{1}^{−1}I_{3}^{−1}=I2I4⊗o(DI1I3)^{∗}.

So from the Theorem 1.3.1 it follows thatf∗gbelongs toC_{K/k}(I1I3D, I2I4).

Now we study the multiplication∗on the modulesB_{K/k}.

Proposition 1.7.2. Let f belong to B_{K/k}(I1, I2) and let g belong to
B_{K/k}(I3, I4), then

f∗g∈B_{K/k}(I1I3D, I2I4).

Proof. SinceB_{K/k}(I, J) is a submodule inC_{K/k}(I, J), from Proposition 1.7 it
follows thatf ∗g∈C_{K/k}(I1I3D, I2I4). From Proposition 1.5 we deduce that
f and gbelong tophi(K⊗kK^{G}), g∈φ(K⊗kK^{G}). Sof∗g∈φ(K⊗kK^{G}),
and this implies that

f∗g∈C_{K/k}(I1I3D, I2I4)∩φ(K⊗kK^{G}) =B_{K/k}(I1I3D, I2I4).

Proposition 1.7.3. Let f belong to A_{K/k}(I1, I2) and LET g belong to
A_{K/k}(I3, I4), then

f∗g∈A_{K/k}(I1I3D, I2I4).

Proof. From Proposition 1.7 it follows that

f∗g∈C_{K/k}(I1I3D, I2I4) (10)
since A_{K/k}(I1, I2) and A_{K/k}(I3, I4) are submodules OF C_{K/k}(I1, I2) and
C_{K/k}(I3, I4) respectively.

From the definition of A_{K/k} it follows that f and g belong to k[G]. So the
coefficients off andglie in k, and from Proposition 1.6.1 it follows thatf ∗g
also belongs tok[G]. Then (10) implies that

f∗g∈C_{K/k}(I1I3D, I2I4)∩k[G] =A_{K/k}(I1I3D, I2I4),
and thus the proposition is proved.

§2 Isomorphism of rings of integers of totally wildly ramified extensions of complete discrete

valuation fields with their associated orders.

2.1. LetK/k be a totally wildly ramified Galois extension of a complete dis-
crete valuation field with residue field of characteristicp. LetDbe the different
of the extension and let O_{K} be the ring of integers of the fieldK. From this
moment and up to the end of the paper we will suppose the condition (*) of
the introduction to be fulfilled, i.e., that D = (δ), withδ ∈k. We will write
A_{K/k}(O_{K}), B_{K/k}(O_{K}) instead ofA_{K/k}(O_{K},O_{K}), B_{K/k}(O_{K},O_{K}).

We denote prime elements of the fieldskandKbyπ0andπ respectively, and
their maximal ideals byM_{o} andM.

Proposition 2.1. The modules C_{K/k}(O_{K}), A_{K/k}(O_{K}), B_{K/k}(O_{K}) are o-
algebras with a unit with respect to the multiplication

f^{∆}∗g=δf∗g=δ^{−1}φ(φ^{−1}(δf)φ^{−1}(δg))
(cf. (9)). The unit is given by δ^{−1}tr.

The motivation for the above definition is given by Theorem 2.4.1 below.

Proof. Let f and g belong to C_{K/k}(O_{K}), then according to Proposition 1.7,
the product f∗g maps the different D into the ringO_{K}. It follows that f^{∆}∗g
maps D into D, and so it also maps o into itself since D =δO_{K}. We obtain
that ^{∆}∗defines a multiplication on the each of the modules associated toO_{K}.
Now we consider the elementδ^{−1}tr and prove that it is the unit for the multi-
plication^{∆}∗in each of these modules. It is clear thatδ^{−1}tr mapsO_{K}into itself
and that δ^{−1}tr belongs tok[G], soδ^{−1}tr belongs toA_{K/k}(O_{K}). Besides that,
δ^{−1}tr commutes with all elements ofGand soδ^{−1}tr lies inB_{K/k}(O_{K}).

Let nowf belong toK[G], then f =X

σ

aσσ, aσ∈K, δ^{−1}tr =X

σ

δ^{−1}σ.

So, according to proposition 1.6.1,

f∗(δ^{−1}tr) =X

σ

δ^{−1}aσσ
and we obtain

f^{∆}∗(δ^{−1}tr) =X
aσ=f.

Since A_{K/k}(O_{K}),B_{K/k}(O_{K}) are o-submodules in C_{K/k}(O_{K}), δ^{−1}tr ∈
A_{K/k}(O_{K}), B_{K/k}(O_{K}), δ^{−1}tr is also an identity in A_{K/k}(O_{K}), B_{K/k}(O_{K})
with respect to the multiplication^{∆}∗.

2.2. Let as beforenbe equal to [K:k].

Lemma 2.2.1. Let x be an element of the ring O_{K} whose valuation equals
n−1, i.e.,vK(x) =n−1. Then

vk(trx) =vk(δ).

In particular, there is an element a in O_{K} with vK(a) =n−1and such that
tra=δ.

Proof. LetMandM_{o}be the maximal ideals ofKandkrespectively. From the
definition of the different and surjectivity of the trace operator it follows that
tr(M^{−1}D^{−1}) =M^{−1}

o . Moreover any element ofM^{−1}D^{−1}, that does not belong
to D^{−1}, has a non-integral trace. So the trace of the element z=π^{−1}_{0} δ^{−1}xis
equal to

trz=π_{0}^{−1}δ^{−1}trx=π^{−1}_{0} ε, ε∈o^{∗}.

Thus trx=εδ. Further, if we multiply the elementxbyε^{−1}∈o^{∗}, then we get
the elementa.

Lemma 2.2.2. In the ringO_{K} we can choose a basisa0, a1, . . . , an−2, a, where
ais as in Lemma 2.2.1, and theai for0≤i≤n−2are such thatvK(ai) =i,
and satisfy tr ai= 0.

Proof. The kernel Ker tr(O_{K}) haso-rank equal ton−1. Letx0, . . . , xn−2be an
o-basis of Ker tr(O_{K}). Along with the elementathey form ao-base of the ring
O_{K}. By elementary operations in Ker tr(O_{K}) we can get from x0, . . . , xn−2

a set of elements with pairwise different valuations. Their valuations have to
be less than n−1. Indeed, otherwise by subtracting from the element x0 of
valuationn−1 an elementaof the same valuation multiplied by a coefficient
in o^{∗} we can obtain an element ofM^{n}, which is impossible.

2.3. Letabe an element ofO_{K}with valuation equal ton−1, wheren= [K:k],
and let

tra=δ, (12)

whereδ is a generator of the differentD_{K/k} (cf. Lemma 2.2.1).

Proposition 2.3.1. 1. The moduleA_{K/k}(O_{K})(a) modM^{n} is a subring with
an identity inO_{K} modM^{n} (with standard multiplication).

2. The multiplication^{∆}∗ inA_{K/k}(O_{K})(cf. (11)) induces the standard multipli-
cation in the ringA_{K/k}(O_{K})(a) modM^{n}, i.e.,

f^{∆}∗g(a)≡f(a)g(a) modM^{n}.

Proof. Letf andgbelong toA_{K/k}(O_{K}). Then the preimagesφ^{−1}(δf), φ^{−1}(δg)
with respect to the bijectionφbelong toO_{K}⊗oO_{K}, since

D⊗oD^{−1}=δO_{K}⊗oδ^{−1}O_{K}=O_{K}⊗oO_{K}.
We prove that

φ^{−1}(δa) =x⊗1 +y, (13)

wherex∈O_{K} andy∈O_{K}⊗M.

Indeed, φ^{−1}(δa) = Pai⊗bi. Ifbi ∈M, then ai⊗bi ∈O_{K}⊗M, otherwise
bi=ci+di, whereci∈o,di∈Mand soai⊗bi=ai⊗ci+ai⊗di=cixi⊗1+yi,
wherecixi ∈O_{K}, sinceci∈oandyi∈O_{K}⊗M. So (13) follows.

Similarly

φ^{−1}(δg) =x^{0}⊗1 +y^{0}, (14)
wherex^{0}∈O_{K},y^{0} ∈O_{K}⊗M.

Thus

φ^{−1}(δf)φ^{−1}(δg) = (x⊗1 +y)(x^{0}⊗1 +y^{0})

=xx^{0}⊗1 +y((x^{0}⊗1) +y^{0}) + (x⊗1)y^{0}

=xx^{0}⊗1 +z, wherez∈O_{K}⊗M.

(15)

We consider the action of the element f ∈ A_{K/k}(O_{K}) on the element awith
valuation equal to n−1. From (13) we obtain f = δ^{−1}φ(x⊗1 +y), where
x∈O_{K},y∈O_{K}⊗M. Then from the definition of the mapφwe have:

f(a) =δ^{−1}φ(x⊗1 +y)(a)

=δ^{−1}(φ(x⊗1)(a) +φ(y)(a))

=δ^{−1}xtra+δ^{−1}φ(y)(a).

We show thatδ^{−1}φ(y)(a)∈M^{n},
i.e.,

f(a) =δ^{−1}xtra+z, wherez∈M^{n}. (16)
Indeed, let y be equal to Pai⊗bi, then from the definition of φ we deduce
that

φ(y)(a) =X

aitr(bia).

Moreover trbia∈ DM_{o}, thus δ^{−1}aitr(bia)∈M^{n}, i.e., z =δ^{−1}φ(y)(a)∈ M^{n}
and we obtain (16).

Our assumptions imply that tra=δ, so

f(a)≡x modM^{n}
and similarly

g(a)≡x^{0} modM^{n},
wherex^{0} is the element from (14). Then

f(a)g(a)≡xx^{0} modM^{n}.

On the other hand from the definition of the multiplication f^{∆}∗g (cf. (11)) it
follows that

f^{∆}∗g(a) =δ^{−1}φ(φ^{−1}(δf)φ^{−1}(δg))(a).

Using this and keeping in mind (15) and (16) we obtain
f^{∆}∗g(a) =δ^{−1}φ(xx^{0}⊗1 +z)(a)≡xx^{0} modM^{n}.
So we have the congruence

f^{∆}∗g(a)≡f(a)g(a) modM^{n}.

Also, the elementδ^{−1}tr inA_{K/k}(O_{K}) gives us an identity element in the ring
A_{K/k}(O_{K})(a) modM^{n}, sinceδ^{−1}tra= 1 (cf. (12)).

Remark 2.3.2. For any other element a^{0} in the ring O_{K} with valuation equal
to n−1 we haveA_{K/k}(O_{K})(a^{0})≡εA_{K/k}(O_{K})(a) modM^{n}, ε∈o^{∗}.

Remark 2.3.3. A similar statement also holds for the module B_{K/k}(O_{K})(a)
modM^{n}.

2.4. Now we formulate the statements which we will begin to prove in the next subsection.

We will investigate the following condition:

in the orderA_{K/k}(O_{K})(resp. B_{K/k}(O_{K})) there exists an elementξ such that

ξ(a) =π, (17)

whereπis a prime element of the fieldKandais some element with valuation equal to n−1.

Theorem 2.4.1. If in the ring A_{K/k}(O_{K}) (resp. B_{K/k}(O_{K})) the condition
(17) is fulfilled, then the element ξ generates a “power” basis of A_{K/k}(O_{K})
(B_{K/k}(O_{K})resp.) over owith respect to the multiplication ^{∆}∗ (cf. 11), i.e.,

A_{K/k}(O_{K}) =hξ^{0}, ξ^{1}, . . . , ξ^{n−1}i,
whereξ^{0}=δ^{−1}tris the unit andξ^{i} =ξ^{∆}∗ξ^{i−1}.

Theorem 2.4.2. 1. If for the ring A_{K/k}(O_{K}) (resp. B_{K/k}(O_{K})) the condi-
tion (17) is fulfilled, then the ring O_{K} is a freeA_{K/k}(O_{K})-module (resp. free
B_{K/k}(O_{K})-module).

2. If the ringO_{K} is a free module over the ringA_{K/k}(O_{K})(resp. B_{K/k}(O_{K}))
and if moreover the orderA_{K/k}(O_{K})(resp. B_{K/k}(O_{K})) is indecomposable (i.e
does not contain non-trivial idempotents), then for the ring A_{K/k}(O_{K}) (resp.

B_{K/k}(O_{K})) the condition (17) and so also the assertions of the theorem 2.4.1
are fulfilled.

Remark 2.4.3. If ξ maps some element with valuation equal to n−1 onto an element with valuation equal to 1, then ξ also maps any other element with valuation equal ton−1 onto an element with valuation equal to 1.

Indeed, if a ∈ K, vk(a) =n−1 and vK(ξ(a)) = 1, then any other element
a^{0} ∈O_{K}, for which vK(a^{0}) =n−1, is equal toεa+b, whereε∈o^{∗}, b∈M^{n},
so b =π0b^{0}, where π0 is a prime element in k. Besides that b^{0} ∈O_{K} and we
obtainξ(b) =π0ξ(b^{0}) and this impliesvK(ξ(b))≥n.

We also have vK(ξ(εa)) = vK(εξ(a)) = vK(ε) + 1 = 1 i.e., ε ∈ o^{∗}, and so
vK(ξ(a^{0})) = 1.

Remark 2.4.4. Any element ξ in A_{K/k}(O_{K}) (resp. in B_{K/k}(O_{K})) that fulfills
the condition (17) generates a power base of the o-moduleA_{K/k}(O_{K}) (resp.

B_{K/k}(O_{K})) with respect to the multiplication^{∆}∗.

2.5 Proof of Theorem 2.4.1 and of the first part of Theorem 2.4.2..

We take the element ain the ring O_{K} such thatvK(a) =n−1 and tra=δ
(cf. (12)). By assumption we haveξ(a) =π, whereπis a prime element of the
field K. We check that

ξ^{i}(a)≡π^{i} modM^{n}

K. (18)

Indeed, if i = 0, then for the usual product ξ^{0}(a)δ^{−1}tr(a) = 1. Let further
ξ^{i−1}(a) be equal to π^{i−1} modM^{n}

K, then according to Proposition 2.3.1 we have

π^{i}≡ξ^{i−1}(a)ξ(a)≡(ξ^{i−1∆}∗ξ)(a)≡ξ^{i}(a) modM^{n}

K

and the congruence (18) is proved. This equality implies that ξ^{i}(a), 0≤i ≤
n−1,generateO_{K}, i.e.,O_{K} =oξ^{0}(a)⊕ · · · ⊕oξ^{n−1}(a).Now we show that

A_{K/k}(O_{K}) =hξ^{0}, . . . , ξ^{n−1}i.

If there exists an element η ∈A_{K/k}(O_{K}) that does not belong to ao-module
hξ^{0}, . . . , ξ^{n−1}i, thenη(a) =b∈O_{K}. So we obtain

η(a) =

n−1X

i=0

αiξ^{i}(a), αi∈o,
i.e.,

(η−X

αiξ^{i})(a) = 0 (19)

Now we show that

η=X

αiξ^{i}. (20)

Indeed the spaces kA_{K/k}(O_{K}) andkhξ^{0}, . . . , ξ^{n−1}ihave equal dimensions and
so coincide. It follows from (19) that

η−X

αiξ^{i}∈A_{K/k}(O_{K})⊂kA_{K/k}(O_{K}) =khξ^{0}, . . . , ξ^{n−1}i,
and so

η−X
αiξ^{i}=

n−1X

i=0

α^{0}_{i}ξ^{i},

whereα^{0}_{i}∈k. Sinceα^{0}_{i}∈kand the valuations of the elementsξ^{i}(a), 0≤i≤n−

1 are pairwise non-congruent modn(cf. (16)), the valuations vK((α^{0}_{i}ξ^{i})(a))
are also pairwise non-congruent modn, and so Pα^{0}_{i}ξ^{i}(a)6= 0 if not all the
α^{0}_{i} are equal to 0. This reasoning proves (20).

So we have obtained that the ring O_{K} is a freeA_{K/k}(O_{K})-module:

O_{K} =A_{K/k}(O_{K})(a)

and we have proved Theorem 2.4.1 and the first part of Theorem 2.4.2.

Lemma 2.6. Let x be an element of the ring O_{K} such that trx= 0, then for
any f ∈A_{K/k}(O_{K})andg∈B_{K/k}(O_{K})the following equalities hold:

trf(x) = 0 = trg(x) = 0.

Proof. Iff ∈A_{K/k}(O_{K}), thenf =P

σ∈Gaσσ, aσ∈k, and so trf(x) = tr(X

aσσ(x)) =X

aσtrσ(x) = 0.

If g ∈ B_{K/k}(O_{K}), then trg(x) = g(tr(x)) = g(0) = 0, since g is an o[G]-
homomorphism.

2.7. Proof of necessity in Theorem 2.4.2.

LetO_{K} be a freeA_{K/k}(O_{K})-module and assume the orderA_{K/k}(O_{K}) IS inde-
composable. We prove that in the orderA_{K/k}(O_{K}) there exists an elementξ
that fulfills the condition (17) of 2.4. We take elementsa0, . . . , an−2such that
vK(ai) =iand such that trai= 0 (cf. Lemma 2.2.2 and also [By1]). We take
further an elementawith valuation equal ton−1. Letχ: O_{K} →A_{K/k}(O_{K})
be an isomorphism of o[G]-modules, then

A_{K/k}(O_{K}) =hχ(a0), . . . , χ(a)i^{o}.
Thus, in particular, there existαi, α∈o such that

1 =α0χ(a0) +· · ·+αn−2+αχ(a).

The ringA_{K/k}(O_{K}) is indecomposable by assumption, and so, according to the
Krull-Schmidt Theorem, it is a local ring. We obtain that one of theχ(ai) OR
χ(a) has to be invertible in the ringA_{K/k}(O_{K}). The elementsχ(ai) cannot be
invertible since, according to Lemma 2.6, A_{K/k}(O_{K})(ai) ∈ Ker trO_{K}. Thus
χ(a) is invertible and it follows that A_{K/k}(O_{K})(a) = O_{K}. We obtain that
there exists aξ in A_{K/k}(O_{K}) such thatξ(a) =π. Theorem 2.4.2 is proved.

Remark 2.7. The corresponding reasoning for the ringB_{K/k}(O_{K}) almost lit-
erally repeats the one we used above.

§3 Kummer extensions for formal groups.

The proof of sufficiency in Theorem A.

Starting from this section we assume that the extensionK/k is Abelian.

3.1. We denote the valuation onkbyv0. We also denote byv0the valuation onK that coincides withv0onk.

We suppose that the fieldkfulfills the conditions of§2 and thatFis some formal
group over the ring o (the coefficients of the series F(X, Y) may, generally
speaking, lie, for example, in the ring of integers of some smaller field). On the
maximal idealM_{o}of the ringowe introduce a structure of a formalZ_{p}-module
using the formal groupF by letting forx, y∈M_{o} andα∈Z_{p}

x+

F y=F(x, y), αx= [α](x).

We denote the Z_{p}-module obtained in this way by F(M_{o}). Let T be a finite
torsion subgroup inF(M_{o}) and letn= cardT be the cardinality of the group
T. Obviously,nis a power ofp.

We construct the following series:

P(X) = Y

t∈T

(X−

Ft). (21)

Remark 3.1.1. The constant term of the series P(X) is equal to zero, the
coefficient atX^{n}is invertible ino, and the coefficients at the powers, not equal
to n, belong to the idealM_{o}.

Lemma 3.1.2. Let a be a prime element of the field k and K = k(x) be the extension obtained from k by adjoining the roots of the equation P(X) = a, where the seriesP is as in (21). Then the extension K/k is a totally ramified Abelian extension of degree nand the differentD of the extensionK/k is gen- erated by an element of the base field, i.e., D = (δ), δ ∈k. The ramification jumps of the extension K/kare equal to hl=nv0(tl)−1, tl∈T.

Proof. Using the Weierstrass Preparation Lemma we decompose the series P(X)−ainto a product

P(X)−a=cf(X)ε(X),

where ε(X) ∈ o[[X]]^{∗} is an invertible series (with respect to multiplication),
f(X) is A unitary polynomial,c∈o. Then, according to Remark 3.1.1, f(X)
is an Eisenstein polynomial of degree n. The series P(X)−a has the same
roots (we consider only the roots ofP(X)−awith positive valuation) as the
polynomialf(X), and so there are exactlynroots. It is obvious that ifP(x) =
a, then

P(x+

F τ) =Y

t∈T

(x−

F t+

Fτ) =Y

t∈T

(x−

Ft) =P(x) =a, τ ∈T and so the roots ofP(X)−aandf(X) are exactly the elementsx+

Fτ, τ ∈T.

Thus we proved that all roots off(X) lie in K and are all distinct. It follows that K/k is a Galois extension. We denote the Galois group of the extension K/k by G. Obviously if σ1, σ2 ∈ G, σ1(x) = x+

F t1, σ2(x) = x+

F t2, then σ2(σ1(x)) = σ2(x+

F t1) = σ2(x) +

F t1 = x+

F t2+

F t1 (as F(X, Y) is defined over o, t1 ∈ T). Since the addition +

F is commutative, the extension K/k is Abelian. We also have that Q

σ(x) =a, vk(a) = 1, and so v0(x) = ^{e(K/k)}_{n} .
This implies that the extension K/k is totally ramified and thatx is a prime
element inK. Now we compute the ramification jumps of the extensionK/k.

LetF(X, Y) =X+Y+P

i,j>0aijX^{i}Y^{j} be the formal group law, then we have
x−σl(x) =x−(x+

F tl) =tl+ X

i,j>0

aijx^{i}t^{j}=tlεt,

where εt is a unit of the ring O_{K}. It follows that the ramification jumps
of the extension K/k are equal to nvk(tl)−1. Hence the exponent of the
different is equal to P

tl∈T(hl+ 1) = nP

tl∈Tvk(tl) (cf., for example, [Se], Ch. 4, Proposition 4) and sovk(D)≡0 modn.

3.2. Before beginning the proof of Theorem A we prove that the first condition of Theorem A is equivalent to a weaker one.

Proposition 3.2. Let a belong to o, vk(a) = ns + 1, where 0 ≤ s <

mint∈Tvk(t). Then the extension k(x)/k, where x is A root of the equation P(X) =a, has the same properties as the extensions from the first condition of Theorem A.

Proof. We consider the series

Fs(X, Y) =π_{0}^{−s}F(π_{0}^{s}X, π_{0}^{s}Y).

It is easily seen that Fs also defines a formal group law and the elements of
Ts={π_{0}^{−s}t, t∈T}form some torsion subgroup in the formal moduleFs(M_{o}).

Indeed, if F(X, Y) =PaijX^{i}Y^{j}, then Fs(X, Y) =Pπ^{s(i+j−1)}_{0} aijX^{i}Y^{j}, and
so the coefficients ofFs(X, Y) are integral. Sinceπ^{−s}_{0} X+

Fs

π_{0}^{−s}Y =π_{0}^{−s}(X+

FY), Fs indeed defines an associative and commutative addition. Besides that if u1, u2 ∈ Ts, then u1 +

Fs

u2 = π^{−s}_{0} (π_{0}^{s}u1+

F π_{0}^{s}u2) ∈ Ts and so Ts is indeed a
subgroup inF(M_{o}).

Now we compute the seriesPFs(X) for the formal groupFs. We obtain:

PFs(X) =Y

t∈T

(X −

Fs

π_{0}^{−s}t) =Y

t

(π_{0}^{−s}(π^{s}_{0}X)−

Fs

π_{0}^{−s}t)

=Y

t

π^{−s}_{0} (π_{0}^{s}X−

F t) =π^{−sn}_{0} PF(π_{0}^{s}X).

Thus the equation Pf(X) =a is equivalent to PFs(π^{−s}_{0} X) =aπ_{0}^{−sn}. Besides
that v(π^{−sn}_{0} a) =v(a)−sn, i.e., π_{0}^{−sn}ais a prime element ink.

Now it remains to note that a root of the equationPf(X) =acan be obtained
by multiplication of a root of the equationPFs(Y) =aπ^{−sn}_{0} byπ^{s}_{0}, and so the
extensions obtained by adjoining the roots of these equations coincide.

3.3 The proof of Theorem A:1 =⇒ 2.

Let K/k be an extension obtained by adjoining the roots of the equation
P(X) = π0. So K =k(x) for some root x. We consider the maximal ideal
M_{⊗} of the tensor product O_{K}⊗O_{K}. Obviously M_{⊗} =O_{K}⊗M+M⊗O_{K}.
We also have M^{i}

⊗ = Pi

j=0M^{j} ⊗M^{i−j}, i > 0, and so it is easily seen that

∩i>0M^{i}_{⊗}= 0. It follows that we can introduceZ_{p}-module structureF(M_{⊗}) on
M_{⊗}. We consider the element

α=x⊗1−

F 1⊗x∈M_{⊗} .
We can defineξ in the following way:

ξ=δ^{−1}φ(α). (22)

We check the following properties of the elementξ:

1. ξ∈A_{K/k}(O_{K})

2. Ify belongs toO_{K} andv(y) =n−1, thenξ(y) is a prime element inK.

For (1):

LetX−

F Y be equal toP

bijX^{i}Y^{j}, then the equalities

φ(α) =X

bijφ(x^{i}⊗1^{i})∗φ(1^{j}⊗x^{j})

= X

σ∈G, i,j

bijx^{i}σ(x^{j})σ =X
(x−

Fσx)σ=X

tσσ ∈k[G]

follow from the definitions ofφand∗. It also follows from Theorem 1.3.1 that
ξ belongs toC_{K/k}(O_{K}). Thusξ ∈k[G]∩C_{K/k}(O_{K}) =A_{K/k}(O_{K}).

For (2):

It easily seen that

(x⊗1)−

F (1⊗x) =x⊗1 +y, (23)

wherey∈M⊗M. Indeed, x⊗1−

F1⊗x=x⊗1 + X

i≥1,j≥0

bijx^{i}⊗x^{j}.

Sincebij∈o,bijx^{i}⊗x^{j} ∈O_{K}⊗Mand we obtain (23).

We can assume that tr(xa) =δ(cf. Remark 2.4.3).

Then, as in Proposition 2.3.1, we have

ξ(a) =δ^{−1}φ(x⊗1 +y)(a)≡x modM^{n}.
Sincev(x) = 1, we have proved the property 2.

3.4. Now we construct explicitly a basis of an associated order for extensions
that fulfill the condition 1 of Theorem A. We have proved above that we can
take the element ξ to be equal to δ^{−1}Ptσσ. Then it follows from Theorem
2.4.1 that

A_{K/k}(O_{K}) =hδ^{−1}X

σ

t^{i}_{σ}σ, i= 0, . . . , n−1i.

Now suppose that K is generated by a root of the equationP(X) =b, where
vk(b) =sn+ 1, s <minv0(tl). In 3.2 we proved thatK may be generated by
a root of the equationPs(X) =bπ_{0}^{−sn}, and so it follows in this case that

A_{K/k}(O_{K}) =hδ^{−1}π^{−si}_{0} t^{i}_{σ}σ, i= 0, . . . , n−1i.